Let $R$ be the region enclosed by the $x$ -axis, the line $x=1$, the line $x=2$, and the curve $y=x^2$. $y$ $x$ ${y=x^2}$ $ 1$ $ 2$ $ R$ A solid is generated by rotating $R$ about the $x$ -axis. What is the volume of the solid? Give an exact answer in terms of $\pi$.
Solution: Let's imagine the solid is made out of many thin slices. $y$ $x$ ${y=x^2}$ Each slice is a cylinder. Let the thickness of each slice be $dx$ and let the radius of the base, as a function of $x$, be $r(x)$. Then, the volume of each slice is $\pi [r(x)]^2\,dx$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [r(x)]^2\,dx$ This is called the disc method. What we now need is to figure out the expression of $r(x)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=x^2}$ $ 1$ $ 2$ $r$ The radius is equal to the distance between the curve $y=x^2$ and the $x$ -axis. In other words, for any $x$ -value, $r(x)=x^2}$. Now we can find an expression for the area of the cylinder's base: $\begin{aligned} &\phantom{=}\pi [r(x)}]^2 \\\\ &=\pi\left(x^2}\right)^2 \\\\ &=\pi x^4 \end{aligned}$ The leftmost endpoint of $R$ is at $x=1$ and the rightmost endpoint is at $x=2$. So the interval of integration is $[1,2]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_1^2 \left(\pi x^4\right)dx \\\\ &=\pi\int_1^2 x^4dx \end{aligned}$ Let's evaluate the integral. $\pi \int^2_1 x^4\, dx=\dfrac{31\pi}{5}$ In conclusion, the volume of the solid is $\dfrac{31\pi}{5}$.